A finite abelian group is the internal direct product of its primary components

Given: The objects and hypotheses in the statement.

Let $G$ be finite abelian and write $|G|=p^a m$ with $p\nmid m$. Then $G(p)$ is a subgroup of order $p^a$. It is the unique subgroup of $G$ having that order. In particular, if $p\nmid |G|$, then $G(p)=\{e\}$. (A p-primary component has the full p-power order and is the unique subgroup of that order).

Let $N_0,\ldots,N_{r-1}\trianglelefteq G$. The following are equivalent: the $N_i$ form an internal direct product of $G$; every $g\in G$ has a unique expression $g=n_0\cdots n_{r-1}$ with $n_i\in N_i$; and the multiplication map $\mu:\prod_{i<r}N_i\to G$ is an isomorphism. These statements include the empty family and the one-factor case. (Internal direct products are external direct products, equivalently every element has a unique factorisation).

Powers are the natural powers of def-group-power and finite products those of def-monoid-finite-product, both taken in the commutative monoid $(\mathbb{Z},\cdot,1)$ of lem-units-of-z. Call $p : r \to \mathbb{Z}$ an injective list of primes when every $p_i$ is prime (def-prime) and $p_i = p_j$ forces $i = j$ (def-injection-surjection-bijection). Let $n \in \mathbb{Z}$ with $n \ge 1$ and let $p : r \to \mathbb{Z}$ be an injective list of primes such that every prime divisor of $n$ equals $p_i$ for some $i < r$. Then, with $v_q$ as in def-p-adic-valuation: 1. $\displaystyle n \;=\; \prod_{i<r} p_i^{\,v_{p_i}(n)}$; 2. $v_q(n) = 0$ for every prime $q$ that is not among $p_0,\dots,p_{r-1}$; 3. the exponents are determined by $n$: if $e : r \to \mathbb{N}$ and $n = \prod_{i<r} p_i^{\,e_i}$, then $e_j = v_{p_j}(n)$ for every $j < r$. Clause 3 needs only injectivity of the list, not the covering hypothesis. (For $n \ge 1$ and any injective list $p : r \to \mathbb{Z}$ of primes containing every prime divisor of $n$, one has $n = \prod_{i<r} p_i^{\,v_{p_i}(n)}$; the exponents are determined by $n$, and $v_q(n) = 0$ for every prime $q$ outside the list).

Let $a, b \in \mathbb{Z}$, not both $0$, and put $I \;:=\; \{\, ax + by \;:\; x, y \in \mathbb{Z} \,\} .$ Then $I$ contains a positive element, and its least positive element is $\gcd(a,b)$ (def-common-divisor-and-gcd). In particular there are integers $x_0, y_0$ with $a x_0 + b y_0 \;=\; \gcd(a,b),$ so the equation $ax + by = \gcd(a,b)$ is solvable in $\mathbb{Z}$. (Bézout's identity: for integers $a, b$ not both zero, $\gcd(a,b)$ is the least positive element of $\{\, ax + by : x, y \in \mathbb{Z} \,\}$; in particular $ax + by = \gcd(a,b)$ has an integer solution).

If $G$ and $H$ are finite groups, then their external direct product is finite and has order $|G\times H|=|G|\,|H|$. (For finite groups $G$ and $H$, $|G\times H|=|G|\,|H|$).

Let $G$ be a finite group and $H\le G$. Then $|G|=[G:H]\,|H|.$ Consequently, under the canonical embedding $\iota:\mathbb N\to\mathbb Z$, $|H|$ divides $|G|$. (Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$).